\documentclass[reqno]{amsart}
\usepackage{hyperref}
\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 85, pp. 1--12.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu (login: ftp)}
\thanks{\copyright 2008 Texas State University - San Marcos.}
\vspace{9mm}}
\begin{document}
\title[\hfilneg EJDE-2008/85\hfil Complete classification and stability]
{Complete classification and stability of equilibria in a
delayed ring network}
\author[, X. Lu, S. Guo\hfil EJDE-2008/85\hfilneg]
{Xuwen Lu, Shangjiang Guo} % not in alphabetical order
\address{Xuwen Lu \newline
College of Mathematics and Econometrics,
Hunan University, Changsha, Hunan 410082, China}
\email{luxuwen1@163.com}
\address{Shangjiang Guo \newline
College of Mathematics and Econometrics,
Hunan University, Changsha, Hunan 410082, China}
\email{shangjguo@hnu.cn}
\thanks{Submitted April 16, 2008. Published June 9, 2008.}
\subjclass[2000]{34K18, 92B20}
\keywords{Delay; neural network; equilibrium; stability}
\begin{abstract}
In this paper, we consider a neural network model consisting of
four neurons with delayed self and nearest-neighbor connections.
We provide a full classification of all equilibria and their
stability in the connection weighter parameter space. Such a
classification is essential for the description of spatio-temporal
patterns of the model system and for the applications to dynamic
memory storage and retrieval.
\end{abstract}
\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{remark}[theorem]{Remark}
\section{Introduction}
Golubitsky \textit{et al} \cite{Golubitsky1} showed
that systems with symmetry can lead to many interesting patterns
of oscillation, which are predictable based on the theory of
equivariant bifurcations. In a series of papers
\cite{Krawcewicz1,Krawcewicz2,Wu}, the theory of
equivariant Hopf bifurcation has been extended to systems with time delays
(functional differential equations). It should be noted that this
theory predicts the possible patterns of oscillation in a system
solely on the symmetry structure of the system. To understand
which patterns occur in a particular system and whether they are
stable, one needs to consider a specific model for a system.
With this in mind, there is interest in applying these
results to models related to the Hopfield-Cohen-Grossberg neural
networks \cite{Cohen,Grossberg,Hopfield1,Hopfield2}, and to models with time
delays \cite{Marcus1,Marcus2}. Such models make an ideal test bed
for this theory, as the models for the individual elements are
quite simple (one variable for each element); yet with the
introduction of time delays the behaviour can be quite complex.
The focus of this work is on networks with a ring structure
with nearest-neighbour (bi-directional) coupling between the
elements. This leads to a system with $\mathbb{D}_n$ symmetry;
i.e., a system which has the symmetries of a polygon with $n$ sides
of equal length. Most of these studies have concerned lower
dimensional systems \cite{Campbell2,Guo3,Ncube} and/or systems
with a single time delay \cite{Guo1,Guo2}. There is also work on
Hopfield-Cohen-Grossberg networks with a ring structure and
uni-directional coupling \cite{Campbell1,Campbell3}.
Recurrent neural networks have found successful applications in
many areas such as associative memory, image processing, signal
processing, and optimization. The qualitative study of such
networks have been a subject of current interest and has benefited
very much from the desired neural network applications. For
example, the global asymptotical stability of a unique equilibrium
is necessary for optimization, but the coexistence of multiple
stable equilibria is critical for a network's capacity to store
and retrieve memories stored as stable equilibria. When designing
an associative memory neural network, we should make as many
stable equilibrium states as possible to provide a memory system
with large information capacity, an attractive region of each
stable equilibrium as large as possible to provide the robustness
and fault tolerance for information processing.
In this paper, we consider the locations and stability of all
equilibria existing in a ring network modelled by the
system of delay differential equations
\begin{equation}\label{eq}
\dot{u}_{i}(t)=-
u_i(t)+\alpha f(u_{i}(t-\tau))+
\beta[f(u_{i-1}(t-\tau))+f(u_{i+1}(t-\tau))],
\end{equation}
with $i$ mod 4, the strength of the self and nearest-neighbour
coupling are denoted by $\alpha$ and $\beta$, and respectively,
$\tau$ denote the corresponding delay. In addition, $f$ is
adequately smooth, and satisfies the following normalization,
monotonicity, concavity, and boundedness conditions:
\begin{itemize}
\item[(C1)] $f(0)=f''(0)=0$, $f '(0)= 1$, and $f'(x)>0$ for all
$x \in \mathbb{R}$;
\item[(C2)] $x f ''(x) < 0$ for all $x\neq 0$;
\item[(C3)] There exists $M>0$ such that $|f(x)|0$, it will cause nonlinear oscillation (see Marcus
and Westervelt~\cite{Marcus2}) and desynchronization induced by
the large scale of networks (see Chen \textit {et
al}~\cite{Chen}). For instance, $8$ branches of periodic solutions
may bifurcate simultaneously from the trivial equilibrium. For
detail, we refer to Guo and Huang~\cite{Guo1,Guo2}. To see how
these periodic solutions and equilibria are connected together to
form the global attractor, it is again critical to have full
classification of the equilibrium structure. Unfortunately, this
issue has not been addressed in the literature.
In this paper, the discussion of synchronous equillibria is only
a small part of the paper: We clearly state in Theorems
\ref{thm21} and \ref{thm22}
that this is only a triviality that is stated for completeness,
which can be generalized to a ring network of $n$ neurons, and of
course available coincides with the paper by Wu et al~(\cite{Wu1})
is for 3 neurons in this aspect. The main result of this paper is
another one (see Theorems \ref{thm23}--\ref{thm27}) and we describe the branching
pattern of those equillibria that can not be found from the paper
by Wu et al~(\cite{Wu1}). More importantly, we consider the
stabilities of these equillibria of different patterns. In the
paper by Wu et al~(\cite{Wu1}), they only consider the stabilities
of synchronous equillibria. The outline of the paper is as
follows. In Section 2 we will completely locate all the equilibria
of system \eqref{eq} and find the conditions ensuring the
existence of equilibria with different patterns. In Section 3 we
will discuss the stability and un-stability of the equilibria in
Section 2.
\section{Patterns of Equilibria}
First of all, we consider the patterns of the equilibria of system
\eqref{eq}. To study the steady state bifurcations of the trivial
solution of \eqref{eq}, one must consider the characteristic
equation of the linearization of \eqref{eq} about the trivial
solution. The characteristic matrix of the linearization of
\eqref{eq} at this equilibrium 0 is:
$$
\mathcal{M}_{4}(0,\lambda) =(\lambda + 1 - \alpha e^{-\lambda\tau})
I - \beta e^{-\lambda\tau} M .
$$
Let
$$
\chi = e^{i\frac{\pi}{2}} ,\quad
v_{j}=(1,\chi^{j},\chi^{2j},\chi^{3j})^{T},
\quad (j = 0 , 1 , 2 , 3 ).
$$
Since
\begin{gather*}
\chi^{4j} = e^{2\pi j i} = 1,\\
\chi^{3j} = e^{\frac{3}{2}\pi j i} = e^{-\frac{\pi j}{2}i} = \chi^{-j},\\
\chi^{2j} = e^{\pi j i} = e^{(2\pi-\pi) j i} = chi^{-2j},
\end{gather*}
it follows that
$$
M v^{j} = \begin{bmatrix}
\chi^{j}+\chi^{3j}\\ 1+\chi^{2j}\\
\chi^{j}+\chi^{3j}\\ 1+\chi^{2j}
\end{bmatrix} = (\chi^{j}+\chi^{-j}) v_{j}\,.
$$
Hence,
\begin{align*}
\mathcal{M}_{4}(0,\lambda) v_{j}
&= (\lambda + 1 -\alpha
e^{-\lambda\tau} - \beta e^{-\lambda\tau}(\chi^{j}+\chi^{-j})) v_{j}\\
&= (\lambda + 1 -\alpha e^{-\lambda\tau} - 2\beta
e^{-\lambda\tau} \cos\frac{2\pi j}{4})v_{j} .
\end{align*}
and the characteristic equation is
\begin{align*}
\det\mathcal{M}_{4}(0,\lambda)
&= \prod{j=0}^{3}(\lambda + 1
-\alpha e^{-\lambda\tau} - 2\beta e^{-\lambda\tau} \cos\frac{2\pi
j}{4})\\
&= (a-2b)\prod{j=1}^{3}(a-2b\cos\frac{2\pi j}{4}) = 0
\end{align*}
where $a = \lambda + 1 - \alpha e^{-\lambda\tau}$ and
$b = \beta e^{-\lambda\tau}$. Thus, we can rewrite
\begin{align*}
&\det\mathcal{M}_{4}(0,\lambda)\\
&=(\lambda + 1 -\alpha
e^{-\lambda\tau})^2(\lambda + 1 -\alpha e^{-\lambda\tau} - 2\beta
e^{-\lambda\tau})(\lambda + 1 -\alpha e^{-\lambda\tau} + 2\beta
e^{-\lambda\tau})\,.
\end{align*}
Clearly the characteristic equation has a simple zero if either
$1- \alpha -2\beta = 0$ or $1 - \alpha + 2\beta = 0$, and a
double zero root if $1 - \alpha = 0$.
Thus $\alpha+2\beta=1$ and $\alpha-2\beta=1$ are potential
standard steady state bifurcation curves,while the line $\alpha=1$
is a potential equivariant state bifurcation line. Since the
nonlinearities in the model \eqref{eq} are odd functions these
will be pitchfork bifurcations.
The simplest nontrivial equilibria are of the form
$(x^{\ast},x^{\ast},x^{\ast},x^{\ast})$ where $x^{\ast}$
satisfies:
\begin{equation}\label{eq3}
-x^{\ast} + (\alpha+2\beta) f (x^{\ast}) = 0
\end{equation}
We call these \textit{synchronous} equilibria since all the four
components are the same.
\begin{theorem}\label{thm21}
If $\alpha+2\beta>1$ then there are at least two nontrivial
synchronous equilibria
$(x^{\ast},x^{\ast},x^{\ast},x^{\ast})$ and
$(-x^{\ast},-x^{\ast},-x^{\ast},-x^{\ast})$. If $\alpha+2\beta<1$,
there are no nontrivial synchronous equilibria.
\end{theorem}
\begin{proof}
Let $F(x)=-x+(\alpha +2\beta) f(x)$ .
It follows that $F$ is odd, $F(0)=0$, $F '(x)=-1+(\alpha +2\beta)
f'(x)$ and $F''(x)=(\alpha +2\beta) f''(x)$.
Let $\alpha+2\beta<1$. Then $F'(0)=-1+\alpha+2\beta< 0$. Since
$f''(x)<0$ for $x>0$ then $f'(x)0$, and hence $F'(x)$ is strictly decreasing in $x\in
[0,+\infty)$. So $F'(x)0$. Thus $F$ is strictly
decreasing for $x>0$ and so $F(x)0$, then
$F(x)$ has no zero in $(0,+\infty)$. Hence system \eqref{eq} has
no nontrivial equilibria.
On the other hand, let $\alpha+2\beta>1$, then $F'(0)>0$ and
$\lim_{x\to +\infty}F(x)=\lim_{x\to
+\infty}(-x+(\alpha+2\beta)f(x))$. It follows from ($C3$) that
$\lim_{x\to +\infty}F(x)<0$. Thus there is an
$x^*>0$ such that $F(x^*)=0$. By symmetry $F(-x^{\ast})=0$ .
\end{proof}
This result about the existence of synchronous equilibria allows
us to make a conclusion about bifurcation of the trivial solution.
\begin{theorem}\label{thm22}
There is a supercritical pitchfork
bifurcation of the trivial solution of system \eqref{eq} at
$\alpha+2\beta=1$, leading to two branches of synchronous
equilibria given by $(\pm x^{\ast},\pm x^{\ast},\pm x^{\ast},\pm
x^{\ast})$ where $x^{\ast}$ satisfies \eqref{eq3}.
\end{theorem}
\begin{proof} It is clear from the characteristic
equation (2) that the trivial solution gains a real eigenvalue
with positive real part as $\alpha+2\beta$ increases and passes
through 1. It follows from Theorem \ref{thm21} that the
synchronous equilibria $(\pm x^{\ast},\pm x^{\ast},\pm
x^{\ast},\pm x^{\ast})$ exist only for $\alpha+2\beta>1$ .
Finally, it can be shown that $x^{\ast}\to 0$ as
$\alpha+2\beta\to 1^{+}$, the result follows.
\end{proof}
As we know, system \eqref{eq} is $\mathbb{D}_4$-equivariant.
Moreover, since the function $f$ is odd, then system \eqref{eq} is
$\mathbb{D}_4\oplus\mathbb{Z}_2$-equivariant. Define the action of
some generators of $\mathbb{D}_4$ on $\mathbb{R}^4$:
\begin{displaymath}
(\rho x)_i=x_{i+1},\quad
(\kappa x)_i=x_{2-i},\quad\zeta x=-x \quad\text{for all }x\in
\mathbb{R}^4.
\end{displaymath}
Then, we can classify system
\eqref{eq}'s equilibria into five cases.
Firstly, the equilibria characterized by $\mathbb{D}_4$'s subgroup
$\mathbb{Z}(\rho\kappa)$, take the forms of $(x_1,x_2,x_2,x_1)$ or
$(x_1,x_1,x_2,x_2)$ with $x_1,x_2\in \mathbb{R}$. Here,we only
consider the existence of equilibrium $(x_1,x_2,x_2,x_1)$ because
the other can be discussed analogously. By \eqref{eq}, components
$x_1,x_2\in \mathbb{R}$ satisfy
\begin{equation}\label{eq4}
\begin{gathered}
x_1-\alpha f(x_1)=\beta f(x_1)+\beta f(x_2),\\
x_2-\alpha f(x_2)=\beta f(x_1)+\beta f(x_2).
\end{gathered}
\end{equation}
\begin{theorem}\label{thm23}
If $\alpha>1$, there are at least four
equilibria: $(\pm x_1,\pm x_2,\pm x_2,\pm x_1)$ or
$(\pm x_1,\pm x_1,\pm x_2,\pm x_2)$, where $x_1,x_2$ satisfy \eqref{eq4}
and $x_1\neq x_2$.
\end{theorem}
\begin{proof}
Rewriting the equations
\begin{gather*}
-x_1+(\alpha+\beta)f(x_1)+\beta f(x_2)=0,\\
-x_2+(\alpha+\beta)f(x_2)+\beta f(x_1)=0.
\end{gather*}
and solving for $x_2$ gives
$$
x_2=\frac{\alpha+\beta}{\beta}x_1-\frac{\alpha^2+2\alpha\beta}{\beta}f(x_1).
$$
Putting this in the first equation gives an equation for $x_1$
alone.
$$
F(x_1)=-x_1+(\alpha+\beta)f(x_1)+\beta
f(\frac{\alpha+\beta}{\beta}x_1-\frac{\alpha^2+2\alpha\beta}{\beta}f(x_1))
$$
Defining $g(x)=
\frac{\alpha+\beta}{\beta}x-\frac{\alpha^2+2\alpha\beta}{\beta}f(x)$
we can write $F(x)=-x+(\alpha+\beta)f(x)+\beta f(g(x))$ and find
$$
F'(x)=-1+(\alpha+\beta)f'(x)+(\alpha+\beta)f'(g(x))
-(\alpha^2+2\alpha\beta)f'(x)f'(g(x)).
$$
In fact, we have $F(0)=0$,
$\lim_{x\to\infty}F(x)<0$ and
$$
F'(0)=-1+2(\alpha+\beta)-\alpha^2-2\alpha\beta
=(1-\alpha)(\alpha+2\beta-1).
$$
So if $\alpha+2\beta<1$ and $\alpha>1$ then $F'(0)>0$. This,
together with the fact that
$\lim_{x\to+\infty}F(x)=-\infty$ and $F(0)=0$
implies that there exists a positive constant $\xi$ such that
$F(\xi)=0$ and hence $x_1=\xi$ and $x_2=g(\xi)$. In what follows,
we show that $x_1\neq x_2$, that is, $\xi\neq g(\xi)$. In fact, if
$\xi= g(\xi)$ then $\xi=(\alpha+2\beta)f(\xi)$, and so
$f(\xi)\geq\xi$, where the equality holds if and only $\xi=0$, and
contradicts the fact $\xi>0$. Hence $\xi\neq g(\xi)$.
If $\alpha+2\beta=1$ and $\alpha>1$ then $\beta<0$, note that
$g(0)=0$ and $\lim_{x\to+\infty}g(x)=-\infty$, so
there exist a positive constant $\xi_1$ such that $g(\xi_1)=0$,
viz., $(\alpha+\beta)\xi_1=\alpha f(\xi_1)$. It is clear that
$F(\xi_1)=((\alpha+\beta)^2-\alpha)f(\xi_1)/\alpha=\beta^2f(\xi_1)/\alpha>0$.
Again since $\lim_{x\to+\infty}F(x)=-\infty$, then
there exists a point $x=x_0$ such that $F(x_0)=0$ and
$x_0\neq g(x_0)$.
If $\alpha+2\beta>1$ and $\alpha>1$, then $F'(0)<0$. Obviously,
$F(x^*)=0$, where $x^*$ is given in Theorem \ref{thm21} such that
$x^*=(\alpha+2\beta)f(x^*)$, that is, $g(x^*)=x^*$. Note that
\begin{align*}
F'(x^*)&=-1+2(\alpha+\beta)f'(x^*)+(\alpha^2+2\alpha\beta)(f'(x^*))^2\\
&=-(\alpha f'(x^*)-1)((\alpha+2\beta)f'(x^*)-1).
\end{align*}
It is clear that $F'(x^*)\neq0$ if $f'(x^*)\neq1/\alpha$ and
$f'(x^*)\neq1/(\alpha+2\beta)$.
Again since
$x^*=(\alpha+2\beta)f(x^*)$, we have $f'(x^*)<1/(\alpha+2\beta)$.
We distinguish two cases to discuss whether $f'(x^*)=1/\alpha$.
\textbf{Case (I)}. If $f'(x^*)\neq1/\alpha$, then $F'(x^*)\neq0$.
\textbf{Case (II)}. If $f'(x^*)=1/\alpha$, then $h(x)=\alpha
f(x)-x$ attains the maximal value at the point $x=x^*$, and
$h(x^*)>0$. Note that $\lim_{x\to+\infty}h(x)<0$,
so there is a point $x=\bar{x}\neq x^*$ such that $h(\bar{x})=0$.
At the point
$x=\bar{x}$, we have $F(\bar{x})=0$ and $g(\bar{x})\neq\bar{x}$.
From above, there is at least one point $x_0$ in interval
$(0,+\infty)$ such that $F(x_0)=0$ and $g(x_0)\neq x_0$. So there
are at least four equilibria: $(\pm x_1,\pm x_2,\pm x_2,\pm x_1)$
or $(\pm x_1,\pm x_1,\pm x_2,\pm x_2)$, where $x_1,x_2$ satisfy
Eq.\eqref{eq4} and $x_1\neq x_2$.
\end{proof}
Secondly, the equilibria characterized by the subgroup
$\mathbb{Z}(\zeta\kappa)$ of $\mathbb{D}_4\oplus\mathbb{Z}_2$,
take the forms of $(0,x,0,-x)$ or $(x,0,-x,0)$ with $x\in
\mathbb{R}$. Here, we only consider the existence of equilibrium
$(0,x,0,-x)$ because the other can be discussed analogously. By
\eqref{eq}, the component $x\in \mathbb{R}$ satisfies
\begin{equation}\label{eq5}
-x+\alpha f(x) =0.
\end{equation}
\begin{theorem}\label{thm24}
If $\alpha-1>0$, then there are
at least four equilibria: $(0,\pm x,0,\mp x)$ or $(\pm x,0,\mp
x,0)$, where $x$ satisfy \eqref{eq5} and $x\neq0$.
\end{theorem}
\begin{proof} Let $F(x)=-x+\alpha f(x)$, then
$F'(x)=-1+\alpha f'(x)$. Note that $F(0)=0$,
$\lim_{x\to \infty}F(x)<0$ and $F'(0)=-1+\alpha>0$.
Since $F$ is a continuous function it follows that there is an
$x^{*}>0$ such that $F(x^{*})=0$. By symmetry $F(-x^{*})=0$ as
well.
If $\alpha\leq1$, note that $00$. We find
$F(x)$ have no root in interval $(0,+\infty)$, as $F(x)$ is odd
function, we know $F(x)$ have only a root: $x=0$. Then induce a
conflict. So the existing of the four equilibria imply $\alpha>1$.
\end{proof}
Thirdly, the equilibria characterized by the subgroup
$\mathbb{Z}(\zeta\rho\kappa)$ of
$\mathbb{D}_4\bigoplus\mathbb{Z}_2$, take the form of
$(x_1,x_2,-x_2,-x_1)$ or $(x_1,-x_1,-x_2,x_2)$ with $x_1,x_2\in
\mathbb{R}$. Here,we only consider the existence of equilibrium
$(x_1,x_2,-x_2,-x_1)$ because the other can be discussed
analogously. By \eqref{eq}, the components $x_1,x_2\in \mathbb{R}$
satisfy
\begin{equation}\label{eq6}
\begin{gathered}
x_1-\alpha f(x_1)=-\beta f(x_1)+\beta f(x_2),\\
x_2-\alpha f(x_2)=\beta f(x_1)-\beta f(x_2).
\end{gathered}
\end{equation}
\begin{theorem} \label{thm5}
If $\alpha>1$, there are at least four
equilibria: $(\pm x_1,\pm x_2,\mp x_2,\mp
x_1)$ or $(\pm x_1,\mp x_1,\mp x_2,\pm x_2)$, where
$x_1,x_2$ satisfy \eqref{eq6}.
\end{theorem}
\begin{proof} Rewriting the equations
\begin{gather*}
-x_1+(\alpha-\beta)f(x_1)+\beta f(x_2)=0,\\
-x_2+(\alpha-\beta)f(x_2)+\beta f(x_1)=0.
\end{gather*}
and solving for $x_2$ gives
$$
x_2=\frac{\alpha-\beta}{\beta}x_1-\frac{\alpha^2-2\alpha\beta}{\beta}f(x_1).
$$
Putting this in the first equation gives an equation for $x_1$
alone.
$$
F(x_1)=-x_1+(\alpha-\beta)f(x_1)+\beta
f(\frac{\alpha-\beta}{\beta}x_1-\frac{\alpha^2-2\alpha\beta}{\beta}f(x_1))
$$
Defining $g(x)=
\frac{\alpha-\beta}{\beta}x-\frac{\alpha^2-2\alpha\beta}{\beta}f(x)$
and hence $F(x)=-x+(\alpha-\beta)f(x)+\beta f(g(x))$, we have
$$
F'(x)=-1+(\alpha-\beta)f'(x)+(\alpha-\beta)f'(g(x))
-(\alpha^2-2\alpha\beta)f'(x)f'(g(x)).
$$
In fact, we know $F(0)=0$, $\lim_{x\to\infty}F(x)<0$
and
$$
F'(0)=-1+2(\alpha-\beta)-\alpha^2+2\alpha\beta
=(1-\alpha)(\alpha-2\beta-1).
$$
So if $\alpha-2\beta<1$ and $\alpha>1$ then $F'(0)>0$. This,
together with the fact that
$\lim_{x\to+\infty}F(x)=-\infty$ and $F(0)=0$ implies
that there exists a positive constant $\xi$ such that $F(\xi)=0$ and
hence $x_1=\xi$ and $x_2=g(\xi)$. In what follows, we show that
$x_1\neq x_2$, viz., $\xi\neq g(\xi)$. In fact, if $\xi= g(\xi)$
then $\xi=(\alpha-2\beta)f(\xi)$, so $f(\xi)\geq\xi$, where the
equality holds if and only $\xi=0$, and contradicts the fact
$\xi>0$. Hence $\xi\neq g(\xi)$.
If $\alpha-2\beta=1$ and $\alpha>1$ then $g(0)=0$ and
$\lim_{x\to+\infty}g(x)=-\infty$, so there exist a
positive constant $\xi_1$ such that $g(\xi_1)=0$, viz.,
$(\alpha-\beta)\xi_1=\alpha f(\xi_1)$. It is clear that
$F(\xi_1)=((\alpha-\beta)^2-\alpha)f(\xi_1)/\alpha=\beta^2f(\xi_1)/\alpha>0$.
Again since $\lim_{x\to+\infty}F(x)=-\infty$, then
there exists a point $x=x_0$ such that $F(x_0)=0$ and $x_0\neq
g(x_0)$.
If $\alpha-2\beta>1$ and $\alpha>1$, then $F'(0)<0$. Obviously,
$F(x^*)=0$, where $x^*$ is given in Theorem 2.1 such that
$x^*=(\alpha-2\beta)f(x^*)$, viz., $g(x^*)=x^*$. Note that
\begin{align*}
F'(x^*)&=-1+2(\alpha-\beta)f'(x^*)-(\alpha^2-2\alpha\beta)(f'(x^*))^2)\\
&=-(\alpha f'(x^*)-1)((\alpha-2\beta)f'(x^*)-1).
\end{align*}
It is clear that $F'(x^*)\neq0$ if $f'(x^*)\neq1/\alpha$ and
$f'(x^*)\neq1/(\alpha-2\beta)$.
Again since $x^*=(\alpha-2\beta)f(x^*)$, we have
$f'(x^*)<1/(\alpha-2\beta)$. We can distinguish two cases to
discuss whether $f'(x^*)=1/\alpha$.
\textbf{Case (I)}. If $f'(x^*)\neq1/\alpha$, then $F'(x^*)\neq0$.
\textbf{Case (II)}. If $f'(x^*)=1/\alpha$, then $h(x)=\alpha
f(x)-x$ attains the maximal value at the point $x=x^*$, and
$h(x^*)>0$. Note that $\lim_{x\to+\infty}h(x)<0$,
so there is a point $x=\bar{x}\neq x^*$ such that $h(\bar{x})=0$.
At the point
$x=\bar{x}$, we have $F(\bar{x})=0$ and $g(\bar{x})\neq\bar{x}$.
From above, there is at least one point $x_0$ in interval
$(0,+\infty)$ such that $F(x_0)=0$ and $g(x_0)\neq x_0$. So there
are at least four equilibria: $(\pm x_1,\pm x_2,\mp x_2,\mp x_1)$
or $(\pm x_1,\mp x_1,\mp x_2,\pm x_2)$, where $x_1,x_2$ satisfy
\eqref{eq6} and $x_1\neq x_2$.
\end{proof}
Fourthly, the equilibria $(x,-x,x,-x)$ with $x\in\mathbb{R}$. By
\eqref{eq}, components $x\in \mathbb{R}$ satisfies
\begin{equation}\label{eq7}
-x+(\alpha-2\beta)f(x) =0.
\end{equation}
By using a similar arguments as the proof of Theorem \ref{thm24},
we have the following result.
\begin{theorem}\label{thm26}
If $\alpha-2\beta>1$, there are at least two equilibria:
$(\pm x,\mp x,\pm x,\mp x)$, where $x$ satisfy \eqref{eq7}.
\end{theorem}
Fifthly, the equilibrium $(x_1,x_2,x_1,x_2)$ with
$x_1,x_2\in\mathbb{R}$. By \eqref{eq}, the components $x_1,x_2\in
\mathbb{R}$ satisfy
\begin{equation}\label{eq8}
\begin{gathered}
-x_{1}+\alpha f(x_{1})+2\beta f(x_{2})=0,\\
-x_{2}+\alpha f(x_{2})+2\beta f(x_{1})=0.
\end{gathered}
\end{equation}
\begin{theorem}\label{thm27}
If $\alpha-2\beta>1$, then there are at least the following
two equilibria: $(\pm x_1,\pm x_2,\pm x_1,\pm x_2)$, where $x_1,x_2$ satisfy
\eqref{eq8} and $x_1\neq x_2$.
\end{theorem}
\begin{proof} From the equations above, we find
$$
x_2=\frac{\alpha}{2\beta}x_1-\frac{\alpha^2-4\beta^2}{2\beta}f(x_1).
$$
Putting this in the first equation gives an equation for $x_1$
alone.
$$
F(x_1)=-x_1+\alpha f(x_1)+2\beta
f(\frac{\alpha}{2\beta}x_1-\frac{\alpha^2-4\beta^2}{2\beta}f(x_1))
$$
Defining $g(x)=
\frac{\alpha}{2\beta}x-\frac{\alpha^2-4\beta^2}{2\beta}f(x)$ we can
write $F(x)=-x+\alpha f(x)+2\beta f(g(x))$ and find
$$
F'(x)=-1+\alpha f'(x)+\alpha
f'(g(x))-(\alpha^2-4\beta^2)f'(x)f'(g(x)).
$$
In fact, we know $F(0)=0$, $\lim_{x\to\infty}F(x)<0$
and
$$
F'(0)=-1+2\alpha-\alpha^2+4\beta^2
=-(\alpha-2\beta-1)(\alpha+2\beta-1).
$$
So if $\alpha+2\beta<1$ and $\alpha-2\beta>1$ then $F'(0)>0$. This,
together with the fact that
$\lim_{x\to+\infty}F(x)=-\infty$ and $F(0)=0$ implies
that there exists a positive constant $\xi$ such that $F(\xi)=0$ and
hence $x_1=\xi$ and $x_2=g(\xi)$. In what follows, we show that
$x_1\neq x_2$, viz., $\xi\neq g(\xi)$. In fact, if $\xi= g(\xi)$
then $\xi=(\alpha+2\beta)f(\xi)$, so $f(\xi)\geq\xi$, where the
equality holds if and only $\xi=0$, and contradicts the fact
$\xi>0$. Hence $\xi\neq g(\xi)$.
If $\alpha+2\beta=1$ and $\alpha-2\beta>1$ then $\beta<0$.
Note that
$g(0)=0$ and $\lim_{x\to+\infty}g(x)=-\infty$, so
there exist a positive constant $\xi_1$ such that $g(\xi_1)=0$,
viz., $\alpha \xi_1=(\alpha-x\beta)f(\xi_1)$. It is clear that
$F(\xi_1)=(\alpha^2-\alpha+2\beta)f(\xi_1)/\alpha=4\beta^2f(\xi_1)/\alpha>0$.
Again since $\lim_{x\to+\infty}F(x)=-\infty$, then
there exists a point $x=x_0$ such that $F(x_0)=0$ and $x_0\neq
g(x_0)$.
If $\alpha+2\beta>1$ and $\alpha-2\beta>1$, then $F'(0)<0$.
Obviously, $F(x^*)=0$, where $x^*$ is given in Theorem \ref{thm21}
such that
$x^*=(\alpha+2\beta)f(x^*)$, viz., $g(x^*)=x^*$. Note that
\begin{align*}
F'(x^*)&=-1+2\alpha f'(x^*)-(\alpha^2-4\beta^2)(f'(x^*))^2\\
&=-((\alpha-2\beta) f'(x^*)-1)((\alpha+2\beta)f'(x^*)-1).\\
\end{align*}
It is clear that $F'(x^*)\neq0$ if $f'(x^*)\neq1/(\alpha-2\beta)$
and $f'(x^*)\neq1/(\alpha+2\beta)$.
Again since
$x^*=(\alpha+2\beta)f(x^*)$, we have $f'(x^*)<1/(\alpha+2\beta)$.
We distinguish two cases to discuss whether
$f'(x^*)=1/(\alpha-2\beta)$.
\textbf{Case (I)}. If $f'(x^*)\neq1/(\alpha-2\beta)$, then
$F'(x^*)\neq0$.
\textbf{Case (II)}. If $f'(x^*)=1/(\alpha-2\beta)$,
then $h(x)=(\alpha-2\beta) f(x)-x$
attains the maximal value at the point $x=x^*$, and $h(x^*)>0$.
Note that $\lim_{x\to+\infty}h(x)<0$, so there is a
point $x=\bar{x}\neq x^*$ such that $h(\bar{x})=0$. At the point
$x=\bar{x}$, we have $F(\bar{x})=0$ and $g(\bar{x})\neq\bar{x}$.
From above, there is at least one point $x_0$ in interval
$(0,+\infty)$ such that $F(x_0)=0$ and $g(x_0)\neq x_0$. So there
are at least two equilibria: $(\pm x_1,\pm x_2,\pm x_1,\pm x_2)$,
where $x_1,x_2$ satisfy \eqref{eq8} and $x_1\neq x_2$.
\end{proof}
\section{Linear Stability Analysis}
To investigate the local stability of the system, we need to
consider the characteristic equation of \eqref{eq} associated with
each equilibrium type. To begin, we calculate the linearization
about a generic nontrivial equilibrium
$(x_1^*,x_2^*,x_3^*,x_4^*)$, viz.,
\begin{equation}\label{eq9}
\dot{\mathbf{x}}=-\mathbf{x}+A\mathbf{x}(t-\tau),
\end{equation}
where
$$
A=\begin{pmatrix}
k_{1}\alpha&k_{2}\beta&0&k_{4}\beta\\
k_{1}\beta&k_{2}\alpha&k_{3}\beta&0\\
0&k_{2}\beta&k_{3}\alpha&k_{4}\beta\\
k_{1}\beta&0&k_{3}\beta&k_{4}\alpha
\end{pmatrix}
$$
and $k_{j}=f'(x_j^*)$. The corresponding characteristic equation
can be found by looking for solutions to (\ref{eq9}) of the form
$\mathbf{x}=e^{\lambda t}\mathbf{u}$, where $\lambda \in
\mathbb{C}$ and $\mathbf{u}\in \mathbb{C}^{4}$. Substituting
$\mathbf{x}=e^{\lambda t}\mathbf{u}$ into (\ref{eq9}) yields an
equation for $\lambda$ and
$\mathbf{u}=[u_{1},u_{2},u_{3},u_{4}]^{T}$:
{\footnotesize
\begin{equation}
\begin{bmatrix}
-1+k_{1}\alpha e^{-\lambda\tau}-\lambda&k_{2}\beta e^{-\lambda\tau}&0
&k_{4}\beta e^{-\lambda\tau}\\
k_{1}\beta e^{-\lambda\tau}&-1+k_{2}\alpha e^{-\lambda\tau}-\lambda
&k_{3}\beta e^{-\lambda\tau}&0\\
0&k_{2}\beta e^{-\lambda\tau}&-1+k_{3}\alpha e^{-\lambda\tau}-\lambda
& k_{4}\beta e^{-\lambda\tau}\\k_{1}\beta e^{-\lambda\tau}&0
&k_{3}\beta e^{-\lambda\tau}&-1+k_{4}\alpha e^{-\lambda\tau}-\lambda
\end{bmatrix}
\begin{bmatrix}
u_{1}\\u_{2}\\
u_{3}\\u_{4}
\end{bmatrix}=0.
\end{equation}}
Requiring nontrivial solutions ($\mathbf{u}\neq 0$) gives the
characteristic equation
{\footnotesize
$$
\det\begin{bmatrix}
-1+k_{1}\alpha e^{-\lambda\tau}-\lambda&k_{2}\beta e^{-\lambda\tau}&0
& k_{4}\beta e^{-\lambda\tau}\\k_{1}\beta e^{-\lambda\tau}&-1+k_{2}\alpha
e^{-\lambda\tau}-\lambda&k_{3}\beta e^{-\lambda\tau}&0\\
0&k_{2}\beta e^{-\lambda\tau}&-1+k_{3}\alpha e^{-\lambda\tau}
-\lambda&k_{4}\beta e^{-\lambda\tau}\\
k_{1}\beta e^{-\lambda\tau}&0&k_{3}\beta
e^{-\lambda\tau}&-1+k_{4}\alpha e^{-\lambda\tau}-\lambda
\end{bmatrix}=0.
$$ }
At the equilibrium point $(x^*,x^*,x^*,x^*)$, we have
$k_{j}=k=f'(x^{*})$, $j=1,2,3,4$, thus the characteristic equation
of the system \eqref{eq} becomes
\begin{equation}\label{eq11}
\begin{aligned}
&(-1-\lambda+\alpha k e^{-\lambda\tau})^{2}(-1-\lambda+\alpha k
e^{-\lambda\tau}-2\beta ke^{-\lambda\tau})\\
&\times (-1-\lambda+\alpha k e^{-\lambda\tau}+2\beta
ke^{-\lambda\tau})\\
&=\prod_{j=1}^{4}(\lambda+1-k\alpha
e^{-\lambda\tau}-2k\beta e^{-\lambda\tau}\cos\frac{\pi j}{2})=0.
\end{aligned}
\end{equation}
For the sake of convenience, for each $j=1,2,3,4$, define
$\Delta_j$: $\mathbb{C}\to \mathbb{C}$ by
$$
\Delta_j(\lambda)=\lambda+1-k\alpha e^{-\lambda\tau}-2k\beta
e^{-\lambda\tau}\cos\frac{\pi j}{2}.
$$
Then, (\ref{eq11}) can be rewriten
$$
\prod_{j=1}^{4}\Delta_j(\lambda)=0
$$
\begin{theorem}\label{thm31}
If the parameters satisfy
$|\beta|0.
$$
Note that
$$
\frac{\mathrm{d}R}{\mathrm{d}\upsilon}=1+k(|\alpha|+2|\beta|)\tau
e^{-\upsilon\tau}>0,
$$
then we have $R(\upsilon)>0$ for all $\upsilon\geq0$, and
$R_j(\upsilon,\omega)>0$ for all $\upsilon\geq0$,
$\omega\in\mathbb{R}$.
Now let $\lambda=\upsilon+i\omega$ be an arbitrary root of the
characteristic equation, then
$R_j(\upsilon,\omega)=I_j(\upsilon,\omega)=0$ for some $j\in
\{1,2,3,4\}$. It follows from the previous discussion that
$\upsilon<0$. Therefore all roots of the characteristic equation
have negative real parts, which means that the equilibrium
$(x^*,x^*,x^*,x^*)$ is locally asymptotically stable for any
$\tau\geq 0$.
\end{proof}
\begin{remark} \label{rmk1} \rm
Theorem \ref{thm31} can be applied to uncoupled oscillators.
Namely, the equilibrium
$(x^*,x^*,x^*,x^*)$ is locally asymptotically stable when
$\beta=0$, $|\alpha|1$, or
$k(\alpha-2\beta)>1$, or $k\alpha>1$, where $k=f'(x^*)$.
\end{theorem}
\begin{proof} At the equilibrium point
$(x^*,x^*,x^*,x^*)$, we have
\begin{gather*}
\Delta_1(0)=\Delta_3(0)=1-k\alpha<0\quad\text{when } k\alpha>1,\\
\Delta_2(0)=1-k(\alpha+2\beta)<0\quad\text{when } k(\alpha+2\beta)>1,\\
\Delta_4(0)=1-k(\alpha-2\beta)<0\quad\text{when } k(\alpha-2\beta)>1.
\end{gather*}
On the other hand, $\Delta_j$, $j=1,2,3,4$ are continuous
functions satisfying
$$
\lim_{\lambda\to\infty}\Delta_j=+\infty\quad(j=1,2,3,4).
$$
Thus, under the assumptions of Theorem \ref{thm32}, there exists
some $j\in\{1,2,3,4\}$ such that the function $\Delta_j(\cdot)$
has at least one positive real zero. Namely, (\ref{eq11}) has at
least one positive real root, and hence the equilibrium
$(x^*,x^*,x^*,x^*)$ is unstable.
\end{proof}
For the equilibrium $(0,x,0,-x)$, we have $k_1=k_3=1$,
$k_2=k_4=k=f'(x)$. Then the associated characteristic equation
becomes
\begin{equation}\label{eq14}
\begin{array}{l}
(1+\lambda-\alpha e^{-\lambda\tau})(1+\lambda-k\alpha
e^{-\lambda\tau})\\
\times\big(k(\alpha^2-4\beta^2)e^{-2\lambda\tau}-(1+k)\alpha
e^{-\lambda\tau}(1+\lambda)+(1+\lambda)^2\big)=0.
\end{array}
\end{equation}
\begin{theorem}\label{thm33}
The equilibrium $(0,x,0,-x)$ is
unstable if it exists.
\end{theorem}
\begin{proof} It follows from Theorem \ref{thm24}
that if system \eqref{eq} has a nonzero equilibrium $(0,x,0,x)$
then $\alpha>1$. However, the condition that $\alpha>1$ implies
that the factor $1+\lambda-\alpha e^{-\lambda\tau}$ in equation
(\ref{eq14}) has a root with a positive real part, and hence that
the characteristic equation of the equilibrium $(0,x,0,x)$ has a
root with a positive real part. Therefore, this kind of
equilibrium is always unstable if it exists.
\end{proof}
It follows from Theorem \ref{thm26} that equilibrium $(u,-u,u,-u)$
exists if $\alpha-2\beta>1$. Note that $f'(u)=f'(-u)$, the
characteristic equation evaluated at this equilibrium
$(u,-u,u,-u)$ is same to that of the synchronous equilibria.
Hence, similar arguments to the proof of Theorems \ref{thm31} and
\ref{thm32} leads to the following result.
\begin{theorem} \label{thm34}
Assume that \ref{eq} has an equilibrium of the type
$(u,-u,u,-u)$, let $k=f'(u)$, then for all $\tau\geq0$, (i) the
equilibrium $(u,-u,u,-u)$ is locally asymptotically stable if
$k(|\alpha|+2|\beta|)<1$; (ii) the equilibrium $(u,-u,u,-u)$ is
unstable if either $k(\alpha+2\beta)>1$, or $k(\alpha-2\beta)>1$,
or $k\alpha>1$.
\end{theorem}
For the equilibrium $(x_1,x_2,x_1,x_2)$, we have $k_3=k_1=f'(x_1)$,
$k_4=k_2=f'(x_2)$. Then the characteristic equation becomes:
\begin{equation}
\begin{aligned}
&(1+\lambda-k_1\alpha e^{-\lambda\tau})(1+\lambda-k_2\alpha
e^{-\lambda\tau})\\
&\times(k_1k_2(\alpha^2-4\beta^2)e^{-2\lambda\tau}-(k_1+k_2)\alpha
e^{-\lambda\tau}(1+\lambda)+(1+\lambda)^2)=0.
\end{aligned}
\end{equation}
Using similar arguments as the proof of Theorem \ref{thm32}, we
have the following result.
\begin{theorem} \label{thm35}
The equilibrium $(x_1,x_2,x_1,x_2)$ is
unstable for any $\tau\geq 0$ if either $k_1\alpha>1$ or
$k_2\alpha>1$, where $k_i=f'(x_i)$, $i=1,2$.
\end{theorem}
For the equilibria $(x_1,x_2,x_2,x_1)$ and $(x_1,x_2,-x_2,-x_1)$,
we have $k_4=k_1=f'(\pm x_1)$, $k_3=k_2=f'(\pm x_2)$. Then the
associated characteristic equation becomes
\begin{equation}\label{eq15}
\begin{aligned}
&k_1^2k_2^2\alpha^2(\alpha^2-4\beta^2)e^{-4\lambda\tau}-2k_1k_2(k_1+k_2)
\alpha(\alpha^2-2\beta^2)e^{-3\lambda\tau}(1+\lambda)\\
&+((k_1^2+4k_1k_2+k_2^2)
\alpha^2e^{-2\lambda\tau}-(k_1+k_2)^2\beta^2e^{-2\lambda\tau})(1+\lambda)^2\\
&- 2(k_1+k_2)\alpha e^{-\lambda\tau}(1+\lambda)^3+(1+\lambda)^4=0
\end{aligned}
\end{equation}
However, it is not easy to discuss the distribution of roots of
the transcendental equation (\ref{eq15}). Some relevant results
will be reported later.
\section{Discussion}
In this paper, we investigate the behavior of a neural network
model \eqref{eq} consisting of four neurons with delayed self and
nearest-neighbor connections. We give analytical results on the
existence, patterns, and stability of equilibria of system
\eqref{eq}. In fact, we may analyze the spatio-temporal patterns
of all the bifurcated periodic solutions from the equilibria
presented in this paper.
It turns out that we may encounter the following bifurcations of
periodic solutions (see, for example, \cite{Guo1,Guo2}).
\begin {itemize}
\item[(i)] Mirror-reflecting waves of the form
$x_j(t)=x_{2-j}(t)$, $t\in \mathbb{R}$, $j\,(\mathrm{mod}\, 4)$;
\item[(ii)] Standing waves of the form
$x_j(t)=x_{2-j}(t-\frac{p}{2})$, $t\in \mathbb{R}$, $ j$(mod 4),
where $p>0$ is a period of $x$;
\item[(iii)] Discrete waves of the form $
x_j(t)=x_{j+1}(t\pm \frac{kp}{4})$, $t\in \mathbb{R}$,
$ j(\mod 4)$, where $p>0$ is a period of $x$.
\end {itemize}
Especially, the discrete waves are also called
\emph{synchronous oscillations} (if $k=0\, (\mod 4)$) or
\emph{phase-locked oscillations} (if $k\neq 0\, (\mod 4)$) as each neuron oscillates just like others except not
necessarily in phase with each other. Moreover, by using a similar
arguments as that in \cite{Guo1}, we may expect that the
nontrivial synchronous equilibria will have the same types of
bifurcations as the trivial equilibrium. Namely, the nontrivial
synchronous equilibria given in \ref{thm21} may undergo either a
standard Hopf bifurcation leading to synchronous oscillations
about both of the nontrivial synchronous equilibria, or an
equivariant Hopf bifurcation leading to 10 branches of
oscillations: 4 phase-locked, 4 standing wave and 2 mirror
reflecting. However, it is much more interesting to discuss the
spatio-temporal patterns of periodic solutions bifurcated from the
asynchronous equilibria obtained in Theorem
\ref{thm22}--\ref{thm27}. Some results about this will be reported
later.
\subsection*{Acknowledgements}
This work was partially supported by grants: 10601016 from the
National Natural Science
Foundation of China, [2007]70 from the Program for New
Century Excellent Talents in University of
Education Ministry of China, and 06JJ3001
from the Hunan Provincial Natural Science Foundation.
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\end{document}